Question: The line $y - x \sqrt{3} + 3 = 0$ intersects the parabola $2y^2 = 2x + 3$ at points $A$ and $B.$  Let $P = (\sqrt{3},0).$  Find $|AP - BP|.$
Answer: First, note that $P$ lies on the line $y - x \sqrt{3} + 3 = 0.$

Solving for $x$ in $2y^2 = 2x + 3,$ we get $x = y^2 - \frac{3}{2}.$  Accordingly, let $A = \left( a^2 - \frac{3}{2}, a \right)$ and $B = \left( b^2 - \frac{3}{2}, b \right).$  We can assume that $a < 0$ and $b > 0.$

[asy]
unitsize(1 cm);

pair A, B, P;

real upperparab(real x) {
  return(sqrt(x + 3/2));
}

real lowerparab(real x) {
  return(-sqrt(x + 3/2));
}

A = (0.847467,-1.53214);
B = (2.94997,2.10949);
P = (sqrt(3),0);

draw(graph(upperparab,-3/2,4));
draw(graph(lowerparab,-3/2,4));
draw(interp(A,B,-0.1)--interp(A,B,1.2));

dot("$A$", A, S);
dot("$B$", B, NW);
dot("$P$", P, SE);
[/asy]

Then the slope of $\overline{AB}$ is
\[
\begin{aligned} \sqrt{3} &= \frac{b - a}{(b^2 - \frac{3}{2}) - (a^2 - \frac{3}{2})} \\
&= \frac{b - a}{b^2 - a^2} \\
&= \frac{b - a}{(b - a)(b + a)} \\
& = \frac{1}{a + b} \end{aligned}
\]The difference between the $y$-coordinates of $A$ and $P$ is $a,$ so the difference between the $x$-coordinates of $A$ and $P$ is $\frac{a}{\sqrt{3}}$.  Then
\[AP = \sqrt{a^2 + \left( \frac{a}{\sqrt{3}} \right)^2} = \sqrt{\frac{4}{3} a^2} = -\frac{2}{\sqrt{3}} a.\]Similarly,
\[BP = \frac{2}{\sqrt{3}} b.\]Therefore,
\[|AP - BP| = \frac{2}{\sqrt{3}} (a + b) = \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \boxed{\frac{2}{3}}.\]